Kinetic energy = (1/2) (mass) (speed)²
BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second. So we'll have to
make that conversion.
KE = (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²
= (725.5) · (48 · 1000 · 1 / 3,600)² (kg) · (km·m·hr / hr·km·sec)²
= (725.5) · ( 40/3 )² · ( kg·m² / sec²)
= 128,978 joules (rounded)
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
D = 1/2 g t^2. It works out to 44.1 meters.
➡ 150hrs.
➡ Time = distance × speed
➡ Time = 15*10
➡ Time = 150hrs ans.
Answer:
Magnitude of the force is 4350N
Explanation:
As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration
v² = v₀² + 2 a x
v₀ = 0
a = v² / 2x
a = 62.5²/(2 × 22)
a = 88.78m/s²
the time you need to get this speed
v = v₀ + a t
t = v / a
t = 62.5 / 88.78
t = 0.704s
Let's caculate the magnitude of the force
F = ma
= 49 × 88.78
= 4350.22
≅ 4350N
Magnitude of the force is 4350N
t = 1,025 s
a = 55.43 m / s²
